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Let A be path connected. If A is a subset of $\mathbb{R}$, why is A an interval (whether it is open, closed or half open)?

It looks trivial but I can't explain why:

My approach was: If we consider two points x, y $\in$ A, then there exists a function f : A $\rightarrow$ A. We can "draw a path from x to y in $\mathbb{R}$ and that has to be a subset of $\mathbb{R}$ and is a subset of A which is an interval.

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You should recall the definitions of a path and path connectedness. Then the claim follows immediately from IVT. – Sebastian Bechtel Mar 7 at 13:06
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The only connected subsets of $\Bbb{R}$ are intervals. – Crostul Mar 7 at 13:06
    
What definition for interval do you use? A proof depends on that. – Jeppe Stig Nielsen Mar 7 at 18:24

Hint:

Intermediate value theorem.

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Can you explain why I'm using the intermediate value theorem? I don't follow. – Kagamine Len Mar 7 at 13:14
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@KagamineLen First, write down the definition of path connectedness. Then, take any two points $x,y$ in $A$ and apply this definition to them. Then, use IVT to prove that the interval $[x,y]$ is a subset of $A$. – 5xum Mar 7 at 13:47

Let $A$ be an open non-empty connected subset of $\mathbb{R}$. Suppose it's not an interval. If it's a one-element set, then there's nothing to prove. Therefore, if it's not a one-element entity, one can find $a$ and $b$ such that $a,b \in A$ so that $\exists c \in ]a,b[$ such that $c$ is not in $A$. Then let $A_1 = ]-\infty, c [ \cap A$ and $A_2 = ]c, +\infty [ \cap A$, so that $A$ is the disjoint union of $A_1$ and $A_2$. Conclude

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